Find the area of the region \(x,y): |x^2-1|≤ y≤ 2\.

The curve $y=|x^{2}-1|$ meets $y=2$ where $|x^{2}-1|=2$, i.e. $x^{2}=3$, so $x=\pm\sqrt3$. The region lies between this curve and the line $y=2$, so the area is $\int_{-\sqrt3}^{\sqrt3}\left(2-|x^{2}-1|\right)dx$. By symmetry this is twice the integral over $[0,\sqrt3]$, where $|x^{2}-1|=1-x^{2}$ on $[0,1]$ and $x^{2}-1$ on $[1,\sqrt3]$. We get $2\!\int_0^{\sqrt3}2\,dx=4\sqrt3$ and $2\!\left[\int_0^1(1-x^{2})dx+\int_1^{\sqrt3}(x^{2}-1)dx\right]=2\left[\frac23+\left(\sqrt3-\frac13-\sqrt3+1\right)\right]$, giving $\frac83$ after simplification. The area is $4\sqrt3-\frac83\approx6.9282-2.6667=4.2615$, i.e. $4.26$. Sketching the region is helpful: it is a strip above the W-shaped curve $y=|x^{2}-1|$ and below the horizontal line, so it is symmetric about the $y$-axis. The modulus changes the integrand at $x=\pm1$, and ignoring this kink would give a different, incorrect area. The exact value is $4\sqrt3-\frac83$.